= Solution
Let $p$ be the incident <momentum> and $q_1,q_2$ the outgoing magnitudes in the <laboratory frame>. Transverse <conservation of momentum> gives $(q_1-q_2)\sin\theta=0$. For the nondegenerate symmetric scattering, $q_1=q_2=q$, and the two outgoing <energies> are equal, $\epsilon=(E+E_0)/2$. Then
$$
p=2q\cos\theta,\quad p^2c^2=E^2-E_0^2,\quad q^2c^2=\frac{(E+E_0)^2}{4}-E_0^2.
$$
Combining these <mass shell> relations yields
$$
\boxed{\cos^2\theta=\frac{E+E_0}{E+3E_0},\qquad \tan^2\theta=\frac{2E_0}{E+E_0}.}
$$
Here $E$ includes the incident electron's <rest energy>. As $E\downarrow E_0$ the limiting angle is $45^\circ$, consistent with the nonrelativistic equal-mass <elastic collision>; at large $E/E_0$ the outgoing tracks become forward directed.
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